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Proof step 21 from proof in-rm-v (previous step)

∀ m, lref, v. eq-entry(m, lref) ∈ q0 - v → (∃ eqent0. eqent0.pnum = m ∧ eqent0 ∈ q0) ⊦ eq-entry(m0, none) ∈ q0 - v, ∃ eqent. eqent.pnum = m0 ∧ (eqent = eq-entry(eqent.pnum, none) ∨ eqent ∈ q0)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: eq-entry(m0, none)

Additional information

Used simplifier rules: ⊦ eq-entry(m, lref).pnum = m; (s) Used terms: eq-entry(m0, none)